Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = (int)1e9 + 70;
const long long MOD = 1e9 + 7;
int n;
const int MAXN = 100006;
vector<int> adj[MAXN];
int jump[MAXN][21];
int top10[MAXN][21][10];
int ts[MAXN][21];
int depth[MAXN];
vector<int> order;
void dfs(int u, int p) {
  order.push_back(u);
  depth[u] = depth[p] + 1;
  jump[u][0] = p;
  for (int i = 1; i <= 20; i++) {
    jump[u][i] = jump[jump[u][i - 1]][i - 1];
    int a = 0;
    int b = 0;
    int j = 0;
    for (; j < 10; j++) {
      if (a == ts[u][i - 1] && b == ts[jump[u][i - 1]][i - 1]) {
        break;
      }
      if (a == ts[u][i - 1]) {
        int c = top10[jump[u][i - 1]][i - 1][b];
        b++;
        if (j != 0 && top10[u][i][j - 1] == c) {
          j--;
        } else {
          top10[u][i][j] = c;
        }
      } else if (b == ts[jump[u][i - 1]][i - 1]) {
        int c = top10[u][i - 1][a];
        a++;
        if (j != 0 && top10[u][i][j - 1] == c) {
          j--;
        } else {
          top10[u][i][j] = c;
        }
      } else {
        int c = min(top10[jump[u][i - 1]][i - 1][b], top10[u][i - 1][a]);
        if (c == top10[jump[u][i - 1]][i - 1][b])
          b++;
        else
          a++;
        if (j != 0 && top10[u][i][j - 1] == c) {
          j--;
        } else {
          top10[u][i][j] = c;
        }
      }
    }
    ts[u][i] = j;
  }
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs(v, u);
  }
}
int LCA(int x, int y) {
  if (depth[x] < depth[y]) swap(x, y);
  for (int i = 20; i >= 0; i--) {
    if (depth[jump[x][i]] >= depth[y]) x = jump[x][i];
  }
  if (x == y) return x;
  for (int i = 20; i >= 0; i--) {
    if (jump[x][i] != jump[y][i]) x = jump[x][i], y = jump[y][i];
  }
  return jump[x][0];
}
int main() {
  if (fopen("input.txt", "r")) freopen("input.txt", "r", stdin);
  int m;
  int q;
  cin >> n >> m >> q;
  int root = n / 3 + 1;
  for (int i = 0; i < n - 1; i++) {
    int u;
    int v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int(t);
    scanf("%d", &t);
    if (ts[t][0] == 10) {
      continue;
    }
    top10[t][0][ts[t][0]] = i;
    ts[t][0]++;
  }
  dfs(root, 0);
  for (int qu = 0; qu < q; qu++) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    int c = LCA(u, v);
    vector<int> ans;
    for (int a = 0; a < ts[c][0]; a++) {
      ans.push_back(top10[c][0][a]);
    }
    vector<int> uv;
    uv.push_back(u);
    uv.push_back(v);
    for (int x : uv) {
      for (int i = 20; i >= 0; i--) {
        if (depth[jump[x][i]] < depth[c]) continue;
        for (int a = 0; a < ts[x][i]; a++) {
          ans.push_back(top10[x][i][a]);
        }
        x = jump[x][i];
        sort(ans.begin(), ans.end());
        while (ans.size() > w) {
          ans.pop_back();
        }
      }
    }
    while (ans.size() > w) {
      ans.pop_back();
    }
    cout << ans.size();
    for (int e : ans) {
      cout << " " << e;
    }
    cout << "\n";
  }
  return 0;
}